Can you please expain the 3rd through 6th equations for me Let us first write th

Question

Can you please expain the 3rd through 6th equations for me

Let us first write the resultant force on the box in the frame in which the incline is at rest. Fr = mg sin theta cos theta x - mgsin2 theta y This resultant force is the same in all inertial frames. Hence, acceleration is same in both the frames. ar = g sin theta cos theta x - gsin2 theta y Consider first the situation as seen in the frame in which the incline is at rest. Initial velocity uo = 0 Considering the point of release as the origin, initial position x0 = 0; y0 = 0. Solving Newton's laws we get Work done is just

Explanation / Answer

here ax = g.sin.cos = d2x/dt2 = dVx /dt

  Vx = g.sin.cos . t + C1, where C1 = integration constant

As at t= 0 u = o, given in qts

0 =  g.sin.cos x 0 + C1

C1 = 0

Vx = g.sin.cos . t = dx/dt

x = 1/2 x  g.sin.cos . t2 + C2

putting t= 0, x =0, C2 = 0

x = 1/2 x  g.sin.cos . t2

Similarly, ay = -g.sin2 = d2y/dt2 = dVy/dt

Vy = -g.sin2 . t + C1

putting t= 0, u =0, C1 comes out to be = 0

Vy = -g.sin2 . t = dy/dt

on integrating above equation and putting limits, t= 0, y = 0,

y = - 1/2 x g.sin2 x t2

6th equation:

P.E. = Fx . X + Fy . Y

subsitute the value from above calculated (Fx = m.g.sin cos and Fy = -m.g.sin2)

putting values of X and Y from above

- P.E. = 1/2m. (g.sincos . t)2 + 1/2 .m. (g.sin2 . t)2 (look for value of Vx and Vy , i have solved above)

   1/2m. Vx2 + 1/2.m. Vy2 = 1/2m. V2 = K.E.

as V = (Vx2 + Vy2)

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